A Few Remarks on Bounded Operators on Topological Vector Spaces

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1 Filomat 30:3 (2016), DOI /FIL Z Published by Faculty of Sciences and Mathematics, University of Niš, Serbia Available at: A Few Remarks on Bounded Operators on Topological Vector Spaces Omid Zabeti a, Ljubiša D.R. Kočinac b a Faculty of Mathematics, University of Sistan and Baluchestan, P.O. Box , Zahedan, Iran b University of Niš, Faculty of Sciences and Mathematics, Niš, Serbia Abstract. We give a few observations on different types of bounded operators on a topological vector space X and their relations with compact operators on X. In particular, we investigate when these bounded operators coincide with compact operators. We also consider similar types of bounded bilinear mappings between topological vector spaces. Some properties of tensor product operators between locally convex spaces are established. In the last part of the paper we deal with operators on topological Riesz spaces. 1. Introduction Throughout the paper, all topological vector spaces are over the scalar field K which is either the field R of real numbers or the field C of complex numbers. A neighborhood U of 0 in a topological vector space will be simply called a zero neighborhood. Let X be a topological vector space. A subset A of X is bounded if for each zero neighborhood U in X there is a scalar λ such that A λu. X is said to be locally bounded if there is a bounded neighborhood of 0 X. In the literature one can find two different notions of bounded operators. In [11, 13, 15] there is the definition which is a definition of nb-bounded operators below, while in [7] one can find the definition which is a definition of bb-bounded operators; the following terminology is from [16]. A linear operator T on X is said to be: nb-bounded if there exists a zero neighborhood U X such that T(U) is a bounded subset of X; bb-bounded if T mappings bounded sets into bounded sets. The class of all nb-bounded operators on X is denoted by B n (X) and is equipped with the topology of uniform convergence on some zero neighborhood. A net (S α ) of nb-bounded operators is said to converge to zero uniformly on a zero neighborhood U X if for each zero neighborhood V, there is an α 0 with S α (U) V for each α α 0. The class of all bb-bounded operators on X is denoted by B b (X) and is endowed with the topology of uniform convergence on bounded sets. Note that a net (S α ) of bb-bounded operators converges to zero 2010 Mathematics Subject Classification. Primary 46A32; Secondary 46A03, 46A40, 47B07, 47L05, 57N17 Keywords. Bounded operator, compact operator, central operator, order bounded below operator, bounded bilinear mapping, tensor product of locally convex spaces, Heine-Borel property, topological Riesz space, locally solid Riesz space Received: 12 August 2015; Revised: 20 October 2015; Accepted: 23 October 2015 Communicated by Ekrem Savaş addresses: o.zabeti@gmail.com (Omid Zabeti), lkocinac@gmail.com (Ljubiša D.R. Kočinac)

2 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), uniformly on a bounded set B X if for each zero neighborhood V, there is an index α 0 with S α (B) V for all α α 0. Also, the class of all continuous operators on X is denoted by B c (X) and is assigned to the topology of equicontinuous convergence. A net(s α ) of continuous operators is convergent to the zero operator equicontinuously if for each zero neighborhood V X there exists a zero neighborhood U such that for every ε > 0 there is an α 0 with the property S α (U) εv for any α α 0. It is easy to see that B n (X) B c (X) B b (X). Also note that the above inclusions become equalities when X is locally bounded; see [16, 18] for more details concerning these operators. Recall that a topological vector space X has the Heine-Borel property if every closed and bounded subset of X is compact. By X Y, we mean the algebraic tensor product space. If X and Y are locally convex spaces, then the symbol X π Y will be used for the algebraic tensor product space endowed with the projective tensor product topology. For a review about projective tensor product of locally convex spaces and the related notions, we refer the reader to [15, Chapter III; 5, 6]; for a topological flavour of topological vector spaces and the relevant aspects, one may consult [13, Chapters I, II, III]. Note that the symbol co(a B) denotes the convex hull of A B. Some notions will be given in the course of exposition in the beginning of a section. The paper is organized in the following way. In Section 2 we further investigate bounded linear operators on topological vector spaces. In Section 3 we introduce bounded bilinear mappings between topological vector spaces and consider some their properties. Section 4 is devoted to a topological approach to the notions of central and order bounded below operators defined on a topological Riesz space. With an appropriate topology, we extend some known results for central and order bounded below operators on a Banach lattice, to the topological Riesz space setting. 2. Linear Mappings Let X be a normed space, K(X) be the space of all compact operators on X, and B(X) be the collection of all bounded linear operators. From the equality T(x) T x, it follows that X is a topological B(X)- module, where the module multiplication is given via the formula (T, x) T(x), for each linear operator T and each x X. On the other hand, it is known that K(X) is a closed subspace of B(X). It is also of interest to investigate situations in which K(X) and B(X) are the same. So, it is natural to see if these results can be generalized to ordinary topological vector spaces and to different classes of bounded and compact operators on them. In [16], two different notions for compact operators on a topological vector space have been introduced. A linear operator T on a topological vector space X is said to be: n-compact if there is a zero neighborhood U X for which T(U) is relatively compact (which means that its closure is compact); b-compact if for each bounded set B X, T(B) is relatively compact. We use the notations K n (X) and K b (X) for the set of all n-compact linear operators and the set of all b-compact linear operators on X, respectively. It is easy to see that K n (X) is a two-sided ideal of B n (X) and K b (X) is a right ideal of B b (X). In this section we investigate some relations between bounded linear operators and compact ones. For more details about bounded and compact operators on topological vector spaces and the related notions see [7, 15, 16, 18]. In [18], it has been proved that each class of bounded linear operators on a topological vector space X, with respect to the appropriate topology, forms a topological algebra. In this section we show that X is a topological A-module, where A is one of the topological algebras B n (X), B c (X), and B b (X), respectively, and

3 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), the module multiplication is given via (T, x) T(x), for every linear operator T and every x X. Let us point out that a topological vector space X is a topological A-module, where A is a topological algebra over the same field, provided that the module multiplication is continuous as a mapping from A X, equipped with product topology, into X. Proposition 2.1. The module multiplication in B n (X) is continuous with respect to the topology of uniform convergence on some zero neighborhood. Proof. Let (x α ) be a net in X which is convergent to zero and (T α ) be a net in B n (X) converging to zero uniformly on some zero neighborhood U X. There is an α 0 such that x α U for each α α 0. If V is an arbitrary zero neighborhood in X, Then, there exists an index α 1 with T α (U) V for each α α 1, so that for sufficiently large α, we have T α (x α ) T α (U) V. Proposition 2.2. The module multiplication in B c (X) is continuous with respect to the equicontinuous convergence topology. Proof. Let (x α ) be a net in X which is convergent to zero and (T α ) be a net of continuous operators converging to zero equicontinuously. Suppose V is an arbitrary zero neighborhood in X. There exist a zero neighborhood U X and an index α 0 with T α (U) V for each α α 0. Choose an α 1 such that x α U for every α α 1. Thus, for sufficiently large α, we conclude T α (x α ) T α (U) V. Proposition 2.3. The module multiplication in B b (X) is sequentially continuous with respect to the topology of uniform convergence on bounded sets. Proof. Let (x n ) be a sequence in X which is convergent to zero and (T n ) be a sequence of bb-bounded operators converging to zero uniformly on bounded sets. Note that E = {x n : n N} is bounded in X. Suppose V is an arbitrary zero neighborhood in X. There exists an n 0 such that T n (E) V for any n > n 0, so that we have T n (x n ) T n (E) V. Question 2.4. Is the module multiplication in B b (X) continuous, in general? Note that for a normed space X, K(X) = B(X) if and only if X is finite dimensional. In this step, we consider some situations where a class of compact linear operators coincides with the corresponding class of bounded operators. Proposition 2.5. K b (X) = B b (X) if and only if X has the Heine-Borel property. Proof. K b (X) = B b (X) I K b (X) X has the Heine-Borel property, where I denotes the identity operator on X. Remark 2.6. Note that when X has the Heine-Borel property, then K n (X) = B n (X). Suppose, for a topological vector space X, K n (X) = B n (X). We consider two cases. First, assume X is locally bounded. Then, I B n (X) I K n (X) X is locally compact X is finite dimensional. The second case, when X is not locally bounded. Then, I B n (X) I K n (X) X is not locally compact X is infinite dimensional.

4 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), Let (X n ) be a sequence of topological vector spaces in which, every X n has the Heine-Borel property. Put X = n=1 X n, with the product topology. It is known that X is a topological vector space. In the following, we establish that each X n has the Heine-Borel property if and only if so has X. Theorem 2.7. Let X = n=1 X n, with the product topology; then X has the Heine-Borel property if and only if each X n has this property, as well. Proof. First, assume that each X n has the Heine-Borel property. We claim that if B X is bounded, then there exist bounded subsets B i X i such that B i=1 B i. Put B i = {x X i : (y 1,..., y i 1, x, y i+1,...) B, y j X j }. Each B i is bounded in X i. Let W i be a zero neighborhood in X i and put W = X 1... X i 1 W i X i Since W is a zero neighborhood in X, there exists a positive scalar α such that B αw, so that B i αw i. Also, it is easy to see that B i=1 B i. Therefore, B B i = B i, i=1 i=1 so that we conclude B is also compact, i.e. that X has the Heine-Borel property. For the converse, suppose X has the Heine-Borel property. Choose a bounded set B n X n. Put B = {0}... {0} B n {0}.... It is an easy job to see that B is bounded in X, so that B is compact. By using Tychonoff s theorem, we conclude that B n is compact and this would complete our claim. Collecting results of Theorem 3.5, Proposition 2.5, and Remark 2.6, we have the following. Corollary 2.8. Let (X n ) be a sequence of topological vector spaces, in which, each X n has the Heine-Borel property. Put X = n=1 X n, with the product topology. Then, K b (X) = B b (X) and K n (X) = B n (X). Remark 2.9. Note that when X has the Heine-Borel property, K n (X) need not be equal to K b (X). For example, consider the identity operator on R N. Indeed, it is b-compact but it fails to be n-compact; nevertheless, X has the Heine-Borel property. Compact operators are not closed in the topologies induced by the corresponding class of bounded linear operators. To see this, consider the following examples. Example K n (X) is not a closed subspace of B n (X), in general. Let X be c 00, the space of all real null sequences, with the uniform norm topology. Suppose that T n is the linear operator defined by T n (x 1, x 2,..., x n,...) = (x 1, 1 2 x 2,..., 1 n x n, 0,...). It is easy to see that each T n is n-compact. Also, (T n ) converges uniformly on N (0), the open unit ball of X 1 with center zero, to the linear operator T defined by T(x 1, x 2,...) = (x 1, 1 2 x 2,...). For, if ε > 0 is arbitrary, there is an n 0 N such that 1 n 0 < ε. So, for each n > n 0, (T n T)(N (0) 1 ) N(0) ε. Now, it is not difficult to see that T is nb-bounded but it is not an n-compact linear operator. Example K b (X) fails to be closed in B b (X), in general. Let X be c 0, the space of all vanishing sequences, with the coordinate-wise topology. Let P n be the projection on the first n-components. By using the Tychonoff s theorem, we can conclude that for each n N, P n is b-compact. Also, (P n ) converges uniformly on bounded sets to the identity operator I. We show that I is not b-compact. Suppose that B is the sequence (a n ) defined by a n = (1, 1,..., 1, 0, 0,...) in which 1 is appeared n times. B is a Cauchy sequence in c 0, so that it is bounded. Also note that B = B. Now, if I K b (c 0 ), then B should be compact. Since B is not complete, this is impossible.

5 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), Bilinear Mappings The notion of a jointly continuous bilinear mapping between topological vector spaces has been studied widely, for example, see [13 15] for more information. In particular, when we deal with the normed spaces framework, these mappings carry bounded sets (with respect to the product topology) to bounded sets. On the other hand, tensor products are a fruitful and handy tool in converting a bilinear mapping to a linear operator in any setting; for example, the projective tensor product for normed spaces and the Fremlin projective tensor products for vector lattices and Banach lattices (see [8, 9] for ample information). In a topological vector space setting, we can consider two different non-equivalent ways to define a bounded bilinear mapping. It turns out that these aspects of boundedness are in a sense intermediate notions of a jointly continuous one. On the other hand, different types of bounded linear operators between topological vector spaces and some of their properties have been investigated (see [16, 18]). In this section, by using the concept of projective tensor product between locally convex spaces, we show that, in a sense, different notions of a bounded bilinear mapping coincide with different aspects of a bounded operator. We prove that for two bounded linear operators, the tensor product operator also has the same boundedness property, as well. Definition 3.1. Let X, Y, and Z be topological vector spaces. A bilinear mapping σ : X Y Z is said to be: (i) n-bounded if there exist some zero neighborhoods U X and V Y such that σ(u V) is bounded in Z; (ii) b-bounded if for any bounded sets B 1 X and B 2 Y, σ(b 1 B 2 ) is bounded in Z. We first show that these concepts of bounded bilinear mappings are not equivalent. Example 3.2. Let X = R N be the space of all real sequences with the Tychonoff product topology. Consider the bilinear mapping σ : X X X defined by σ(x, y) = xy where x = (x i ), y = (y i ) and the product is pointwise. It is easily verified that σ is b-bounded; but since X is not locally bounded, it can not be an n-bounded bilinear mapping. It is not difficult to see that every n-bounded bilinear mapping is jointly continuous and every jointly continuous bilinear mapping is b-bounded, so that these concepts of bounded bilinear mappings are related to jointly continuous bilinear mappings. Note that a b-bounded bilinear mapping need not be jointly continuous, even separately continuous; by a separately continuous bilinear mapping, we mean one which is continuous in each of its components. Consider the following example (which is actually originally an exercise from [13, Chapter I, Exercise 13]; we will give a proof for it for the sake of completeness). Example 3.3. Let X be the space C[0, 1], consisting of all real continuous functions on [0, 1]. Suppose τ 1 is the topology generated by the seminorms p x ( f ) = f (x), for each x [0, 1], and τ 2 is the topology induced by the metric defined via the formulae d( f, ) = 1 0 f (x) (x) 1 + f (x) (x) dx. Consider the bilinear mapping σ : (X, τ 1 ) (X, τ 1 ) (X, τ 2 ) defined by σ( f, ) = f. It is easy to show that σ is a b-bounded bilinear mapping. But is it not even separately continuous; for example the mapping = 1 X, the identity operator from (X, τ 1 ) into (X, τ 2 ), is not continuous. To see this, suppose V = { f X : d( f, 0) < 1 2 }. V is a zero neighborhood in (X, τ 2 ). If the identity operator is continuous, there should be a zero neighborhood U (X, τ 1 ) with U V. Therefore, there are {x 1,..., x n } in [0, 1] and ε > 0 such that U = { f X, f (x i ) < ε, i = 1,..., n}.

6 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), For each subinterval [x i, x i+1 ], consider positive reals α i and α i+1 such that x i < α i < α i+1 < x i+1. For an n N, Define, n(x x i ) α i x i, if x i x α i, f i (x) = n, if α i x α i+1, n(x x i+1 ) α i+1 x i+1, if α i+1 x x i+1. Now consider the continuous function f on [0, 1] defined by f i s. Obviously, f U. Put β = n i=1 (α i+1 α i ). We can choose n N and β in such a way that βn n+1 > 1 2. Thus, 1 0 f (x) βn dx > 1 + f (x) n + 1 > 1 2. This completes the claim. In what follows, by using the concept of the projective tensor product of locally convex spaces, we are going to show that these concepts of bounded bilinear mappings are, in fact, the different types of bounded operators defined on a locally convex topological vector space. Recall that if U and V are zero neighborhoods for locally convex spaces X and Y, respectively, then co(u V) is a typical zero neighborhood for the locally convex space X π Y. Proposition 3.4. Let X, Y and Z be locally convex vector spaces and θ : X Y X π Y be the canonical bilinear mapping. If ϕ : X Y Z is an n-bounded bilinear mapping, there exists an nb-bounded operator T : X π Y Z such that T θ = ϕ. Proof. By [15, III.6.1], there is a linear mapping T : X π Y Z such that T θ = ϕ. Therefore, it is enough to show that T is nb-bounded. Since ϕ is n-bounded, there are zero neighborhoods U X and V Y such that ϕ(u V) is bounded in Z. Let W be an arbitrary zero neighborhood in Z. There is r > 0 with ϕ(u V) rw. It is not hard to show that T(U V) rw, so that T(co(U V)) rw. But, by the fact mentioned before this proposition, co(u V) is a zero neighborhood in the space X π Y. This completes the proof. Proposition 3.5. Let X, Y and Z be locally convex vector spaces and θ : X Y X π Y be the canonical bilinear mapping. If ϕ : X Y Z is a b-bounded bilinear mapping, there exists a bb-bounded operator T : X π Y Z such that T θ = ϕ. Proof. As in the proof of the previous theorem, the existence of the linear mapping T : X π Y Z such that T θ = ϕ follows by [15, III.6.1]. We prove that the linear mapping T is bb-bounded. Consider a bounded set B X π Y. There exist bounded sets B 1 X and B 2 Y such that B B 1 B 2. To see this, put B 1 = {x X, y Y, such that x y B}, B 2 = {y Y, x X, such that x y B}. It is not difficult to see that B 1 and B 2 are bounded in X and Y, respectively, and B B 1 B 2. Also, since θ is jointly continuous, B 1 B 2 is also bounded in X π Y. Thus, from the inclusion T(B) T(B 1 B 2 ) = T θ(b 1 B 2 ) = ϕ(b 1 B 2 ) and using the fact that ϕ is a b-bounded bilinear mapping, it follows that T is a bb-bounded linear operator. This concludes the claim and completes the proof of the proposition. Remark 3.6. Note that the similar result for jointly continuous bilinear mappings between locally convex spaces is known and commonly can be found in the contexts concerning topological vector spaces (see for example [15, III.6]).

7 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), We are going now to investigate whether or not the tensor product of two operators preserves different kinds of bounded operators between topological vector spaces. The response is affirmative. Recall that for vector spaces X, Y, Z, and W, and linear operators T : X Y, S : Z W, by the tensor product of T and S, we mean the unique linear operator T S : X Z Y W defined via the formulae (T S)(x z) = T(x) S(z); one may consult [14] for a comprehensive study regarding the tensor product operators. Theorem 3.7. Let X, Y, Z, and W be locally convex spaces, and T : X Y and S : Z W be nb-bounded linear operators. Then the tensor product operator T S : X π Z Y π W is nb-bounded. Proof. Let U X and V Z be two zero neighborhoods such that T(U) and S(V) are bounded subsets of Y and W, respectively. Let O 1 Y and O 2 W be two arbitrary zero neighborhoods. There exist positive reals α and β with T(U) αo 1 and S(V) βo 2. Then (T S)(U V) = T(U) S(V) αβ(o 1 O 2 ) αβco(o 1 O 2 ), so that (T S)(co(U V)) αβco(o 1 O 2 ). This is the desired result. Theorem 3.8. Suppose X, Y, Z, and W are locally convex spaces, and T : X Y and S : Z W are bb-bounded linear operators. Then the tensor product operator T S : X π Z Y π W is also bb-bounded. Proof. Fix a bounded set B X π Z. By the argument used in Proposition 3.5, there are bounded sets B 1 X and B 2 Z with B B 1 B 2. Let O 1 Y and O 2 W be two arbitrary zero neighborhoods. There are positive reals α and β such that T(B 1 ) αo 1 and S(B 2 ) βo 2. Therefore, (T S)(B) (T S)(B 1 B 2 ) = T(B 1 ) S(B 2 ) αβ(o 1 O 2 ) αβco(o 1 O 2 ), hence (T S)(B) αβco(o 1 O 2 ), as required. Theorem 3.9. Suppose X, Y, Z, and W are locally convex spaces, and T : X Y and S : Z W are continuous linear operators. Then the tensor product operator T S : X π Z Y π W is jointly continuous. Proof. Let O 1 Y and O 2 Z be two arbitrary zero neighborhoods. There exist zero neighborhoods U X and V Z such that T(U) O 1 and S(V) O 2. It follows (T S)(U V) = T(U) S(V) (O 1 O 2 ) co(o 1 O 2 ), so that (T S)(co(U V)) co(o 1 O 2 ), as claimed. 4. Operators in Topological Riesz Spaces In this section we give a topological approach to the notions of central operators and order bounded below operators defined on a topological Riesz space. With an appropriate topology, we extend to topological Riesz spaces some known results for these operators on Banach lattices. We recall some concepts and terminology. A Riesz space (or vector lattice) is an ordered real vector space X which is also a lattice. For x X one defines x + = x 0 (the positive part of x), x = ( x) 0 (the negative part of x), and x = x ( x) = x + + x (the absolute value or modulus of x). A subset S of a Riesz space X is said to be solid if y S, x X, and x y imply x S. A topological Riesz space is a Riesz space which is at the same time a (ordered) topological vector space. By a locally solid Riesz space we mean a topological Riesz space with a locally solid topology. A Banach lattice is a Riesz space which is also a Banach space, where the norm is a lattice norm. For more information on topological Riesz spaces and related notions, and also for terminology used in this section, we refer the reader to [1 4, 10].

8 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), Central operators We recall that a linear operator T on a Riesz space X is called central if there exists a positive real number γ such that for each x X, we have T(x) γ x (see [5, 6, 17], where there are also interesting results concerning this class of operators). In [17], Wickstead showed that if X is a Banach lattice, then Z(X), the space of all central operators on X (called the center of X), is a unital Banach algebra with respect to the norm given by T = inf{λ 0 : T λi}, for each T Z(X), where I denotes the identity operator on X. We are going to generalize this result to central operators on a locally solid Riesz space X endowed with the τ-topology: a net (T α ) of central operators converges to zero in the τ-topology if for each ε > 0 there is an α 0 such that T α (x) ε x, for each α α 0 and for each x X. It is easy to see that Z(X) is a unital algebra. We show that Z(X) is in fact a topological algebra. Proposition 4.1. The operations of addition, scalar multiplication and product are continuous in Z(X) with respect to the τ-topology. Proof. Suppose (T α ) and (S α ) are two nets of central operators which are convergent to zero in the τ-topology. Let ε > 0 be given. There are indices α 0 and α 1 such that T α (x) ε 2 x for each α α 0 and x X, and S α (x) ε 2 x for each α α 1 and x X. Choose α 2 with α 2 α 0 and α 2 α 1. Then for each α α 2 and for each x X, we have, (T α + S α )(x) T α (x) + S α (x) ε x. Now, we show the continuity of the scalar multiplication. Suppose (γ α ) is a net of reals which converges to zero. Without loss of generality, we may assume that γ α 1 for each α. Therefore, for each x X we have γ α T α (x) γ α ε x ε x, for all α α 0. For continuity of the product, we have for x X T α (S α (x)) T α S α (x) T α (ε x ) = T α (ε x ) ε 2 x, so that T α (S α (x)) ε 2 x. Note that by [5, 6], for a central operator T on a vector lattice X, the modulus of T, T, exists and satisfies T ( x ) = T(x), for any x X. Proposition 4.2. Suppose (T α ) is a net of central operators on a topological Riesz space X converging (in the τ-topology) to a linear operator T. Then, T is also central. Proof. There is an α 0 such that for each α α 0 and for each x X, we have (T α T)(x) x. Fix an α α 0. There exists a positive real γ α such that T α (x) γ α x. So, we have T(x) T α (x) + x γ α x + x = (γ α + 1) x. Therefore, T is also a central operator. Proposition 4.3. Let X be a complete locally solid Riesz space. τ-topology. Then Z(X) is also complete with respect to the Proof. Let (T α ) be a Cauchy net in Z(X) and V be an arbitrary zero neighborhood in X. Fix x 0 X. Choose δ > 0 such that δ x 0 V. There exists an α 0 such that (T α T β )(x) δ x for each α α 0 and for each β α 0. So, we conclude that (T α T β )(x 0 ) V. This means that (T α (x 0 )) is a Cauchy net in X. Therefore, it is convergent. Put T(x) = lim T α (x). Since this convergence holds in Z(X), by Proposition 4.2, T is also central.

9 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), Collecting the results of Propositions 4.1, 4.2 and 4.3, we have the following Theorem 4.4. Let X be a complete locally solid Riesz space. Then, Z(X) is a complete unital topological algebra. In addition, we have continuity of the lattice operations (defined via formulas of [5, 6]) with respect to the assumed topology, which is proved in the following theorem; in other words (Z(X), τ) is a locally solid Riesz space. Theorem 4.5. The lattice operations of Z(X) are continuous with respect to the assumed topology. Proof. By [5, 6], the supremum and the infimum operations in Z(X) are given by the formulas (T S)(x) = T(x) S(x) and (T S)(x) = T(x) S(x), T, S Z(X), x X +. Let (T α ) and (S α ) be two nets of central operators which are convergent to operators T and S in the τ- topology, respectively. Let ε > 0 be arbitrary. There are some α 0 and α 1 such that for each x X, we have (T α T)(x) ε 2 x for each α α 0 and (S α S)(x) ε 2 x for each α α 1. Pick an α 2 with α 2 α 0 and α 2 α 1. Then for each α α 2, by using the Birkhoff s inequality (for example, see [3]), for each x X + = {y X : y 0}, (T α S α )(x) (T S)(x) = (T α S α )(x) (T α S)(x) + (T α S)(x) (T S)(x) T α (x) S α (x) T α (x) S(x) + T α (x) S(x) T(x) S(x) (T α T)(x) + (S α S)(x) εx. If x is not positive, by using x = x + x, we get (T α S α )(x) (T S)(x) 2ε x for sufficiently large α. Since, the lattice operations in a Riesz space can be obtained via each other, we conclude that all of them are continuous in Z(X) with respect to the τ-topology Order bounded below operators In this subsection we present a new approach to order bounded below operators on a topological Riesz space. The concept of a bounded below operator on a Banach space has been studied widely (for example, see Section 2.1 in [1]), and the norm of the operator was used essentially in its definition. In a Riesz space we have the concepts order and modulus which help us to have a different vision to these operators. On the other hand, when we deal with a topological Riesz space and an operator on it, the concept neighborhoods enables us to consider a topological view for an order bounded below operator. With the topology introduced in the previous subsection we extend some results of [1, Section 2.1] to topological Riesz spaces. A linear operator T on a topological Riesz space X is said to be order bounded below if T(x) γ x for some positive real number γ and each x X. The class of all order bounded below operators on a topological Riesz space X is denoted by O(X). We consider the τ-topology on it. It is easy to see that every order bounded below operator is one-to-one; also, every central order bounded below operator on a locally solid Riesz space has a closed range. In the following lemma we show that if a linear operator is sufficiently close to an order bounded below operator with respect to the τ-topology, then it is also order bounded below. So, the set of all central operators on a topological Riesz space X which are also order bounded below is an open subset of Z(X). Lemma 4.6. If a linear operator S on a topological Riesz space X is sufficiently close to an order bounded below operator T with respect to the τ-topology, then S is also order bounded below.

10 Lj.D.R. Kočinac, O. Zabeti / Filomat 30:3 (2016), Proof. Since T is an order bounded below operator on X there is a positive real number γ such that T(x) γ x for each x X. Choose a linear operator S such that (T S)(x) γ 2 x. Then, for each x X, we have Sx = (S T)(x) + T(x) γ x γ 2 x = γ 2 x, so that S(x) γ 2 x, which completes the proof. Now, we assert the main theorem of this subsection which is an extension of Theorem 2.9 from [1] to a locally solid Riesz space. Theorem 4.7. Let X be a complete locally solid Riesz space and T a continuous order bounded below operator on X. If a net (T α ) of surjective operators converges in the τ-topology to T, then T is also surjective. Proof. Fix a constant γ > 0 such that T(x) 2γ x for each x X. There is an α 0 with (T α T)(x) γ x for each x X and for each α α 0, so that T α (x) γ x. Pick y X. To show that T is surjective, we have to show that there exists x X such that T(x) = y. There is a net {x α } in X such that T α (x α ) = y. From γ x α T α (x α ) = y, it follows that x α 1 γ y. We claim that {x α } is a Cauchy net in X. Let V be an arbitrary zero neighborhood in X. Choose δ > 0 such that δ 1 γ y V. Then for sufficiently large α and β we have 2γ x α x β T(x α ) T(x β ) T(x α ) y + y T(x β ) = T(x α ) T α (x α ) + T β (x β ) T(x β ) δ 2 x α + δ 2 x β δ 1 γ y. So, we conclude that {x α } is really a Cauchy net in X. Suppose that (x α ) converges to x. Thus, it follows that (T(x α )) converges to T(x). On the other hand, from the equality T(x α ) y = T(x α ) T α (x α ) δ x α δ 1 γ y it follows that (T(x α )) converges to y X, hence T(x) = y. This complete the proof. References [1] Y.A. Abramovich, C.D. Aliprantis, An Invitation to Operator Theory, Graduate Studies in Mathematics 50, American Mathematical Society, [2] Y.A. Abramovich, C.D. Aliprantis, I.A. Polyrakis, Some remarks on surjective and bounded below operators, Atti Sem. Mat. Fis. Univ. Modena 44 (1996) [3] C.D. Aliprantis, O. Birkinshaw, Locally Solid Riesz Spaces, Academic Press, New York and London, [4] C.D. Aliprantis, O. Birkinshaw, Positive Operators, Academic Press, New York and London, [5] A. Bigard, K. Keimel, Sur les endomorphismes conversant les polaires d un group réticuléàrchimédien, Boll. Soc. Math. France 97 (1969) [6] P.F. Conard, J.E. Diem, The ring of polar preserving endomorphisms of an Abelian lattice-ordered group, Illinois J. Math. 15 (1971) [7] R.E. Edwards, Functional Analysis. Theory and Applications (Corrected reprint of the 1965 original), Dover Publications, Inc., New York, [8] D.H. Fremlin, Tensor products of Archimedean vector lattices, Amer. J. Math. 94 (1972) [9] D.H. Fremlin, Tensor products of Banach lattices, Math. Ann. 211 (1974) [10] D.H. Fremlin, Topological Riesz Spaces and Measure Theory, Cambridge University Press, 2008 (digitally printed version of the 1974 edition). [11] J. Kelly, I. Namioka, Linear Topological Spaces, D. Van Nostrand Company, Inc., Princeton, [12] Lj.D.R. Kočinac, O. Zabeti, Topological groups of bounded homomorphisms on a topological group, Filomat, this issue. [13] W. Rudin, Functional Analysis (Second edition), McGraw- Hill, New York, [14] R.A. Ryan, Introduction to Tensor Products of Banach Spaces, Monagraphs in Mathematics, Springer-Verlag, London, [15] H.H. Schaefer, Topological Vector Spaces (Second edition), Graduate Texts in Mathematics 3, Springer-Verlag, New York, [16] V.G. Troitsky, Spectral radii of bounded operators on topological vector spaces, Panamer. Math. J. 11:3 (2001) [17] A.W. Wickstead, Representation and duality of multiplication operators on Archimedean Riesz spaces, Compositio Math. 35 (1977) [18] O. Zabeti, Topological algebras of bounded operators on topological vector spaces, J. Adv. Res. Pure Math. 3:1 (2011)